# Finding the value of $\sin {\frac{31 \pi}3}$

The task was to find out the value of $$\sin\frac{31\pi}3$$

This is a example in my book in which following steps are shown: \begin{align} \sin\frac{31\pi}3& =\sin\left(10\pi +\frac\pi3\right)\\ &=\sin\frac\pi3\\ &=\frac{\sqrt3}2 \end{align}

I cannot understand step 3, $=\sin\frac\pi3$

• $$\sin(n\pi+\theta)=\left\{ \begin{array}{l l} \sin\theta & ;\quad \text{if n is even},\\ \\ -\sin\theta & ;\quad \text{if n is odd}. \end{array} \right.$$ May 28, 2014 at 14:34
• I think all answers are good. But i think we can accept only one.Is there any way by which i can accept more than one ? May 28, 2014 at 14:49
• @Tunk-Fey you are right May 28, 2014 at 14:58
• Old math joke: The value of sin? Value==wages and biblically, the wages of sin is death, so the value of sin==death. Try that one on your math teacher. May 28, 2014 at 17:09
• @TechZen i am definetly going to try this on my maths teacher May 28, 2014 at 17:45

The sine function is periodic with period $2\pi$. This means that $\sin\theta = \sin(\theta + 2\pi) \ \forall \ \theta$. One can apply this identity five times to get rid of the $10 \pi$ term.

• What? 8 upvotes? How did that happen!? May 28, 2014 at 21:48
• By the time you have read my last comment, the number 8 will have become obsolete. May 28, 2014 at 22:04
• Really...really...11 upvotes on this question May 29, 2014 at 9:35
• @user3001408 This is answer :) Oct 10, 2014 at 5:56

The sine function is periodic with period $2\pi$. A periodic function is a function that repeats its values in regular intervals or periods. A function is said to be periodic with period $P$, where $P$ being a nonzero constant if we have $$f(x+P) = f(x)$$ for all values of $x$ in the domain. Therefore $$\sin(n\pi+\theta)=\left\{ \begin{array}{l l} \sin\theta & ;\quad \text{if n is even},\\ \\ -\sin\theta & ;\quad \text{if n is odd}. \end{array} \right.$$

• Unfortunately the upvotes do not increase my reputation right now. :( $$\\$$Thanks for the upvoters. :) May 28, 2014 at 14:41
• I think later it will be added up :) May 28, 2014 at 14:54

Multiples of $2\pi$ do not matter because $\sin$ is periodic of period $2\pi$.

• The key fact is that sine is $2\pi$-periodic, and not simply periodic. May 28, 2014 at 14:42
• All answer are good but i am choosing this as it is shortest one May 28, 2014 at 14:59

Think of the unit circle and the definition of $\sin$: A straight line drawn from the origin with the angle between the $x$-axis of $\theta$ intersects the unit circle at $(\cos \theta, \sin \theta)$.

Say you have a point on the unit circle at $(\cos \theta, \sin \theta)$. If the line making the angle $\theta$ from the origin makes a full revolution (revolves around the origin by a measure of $2\pi$), then it will end up in the same position. Thus, $\sin n = \sin (2\pi + n) = \sin (n - 2\pi)$.

You can confirm this by using the angle-sum formula:

\begin{align} \sin(10\pi + \pi/3) & = \sin(10\pi)\cos(\pi/3) + \cos(10 \pi)\sin(\pi/3) \\ \\ & = 0\cdot\frac 12 + 1 \cdot \sin (\pi/3) \\ \\ & = \sin(\pi/3)\\ \\ & = \sqrt 3/2\end{align}